How to change the limits of integration The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integration limits of the double integral after conversion to the polar coordinatesConvergent or Divergent using LimitsHow do I solve a double integral with an absolute value?How are limits of integration changed?Converting limits of integrationHow to set the limits for Jacobian IntegrationLimits of integration in multivariable integrals during change of variablesChange of limits of integrationWhat theorem(s) is(are) used to change between the various improper integrals.how does an integral becoming negative effect limits of integration?
How to pronounce 1ターン?
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
How does ice melt when immersed in water
Empty set is subset of every set? If yes, why that...
Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?
When did F become S in typeography, and why?
University's motivation for having tenure-track positions
Derivation tree not rendering
Difference between "generating set" and free product?
Can a 1st-level character have an ability score above 18?
Can smartphones with the same camera sensor have different image quality?
Does Parliament hold absolute power in the UK?
How do I add random spotting to the same face in cycles?
Do warforged have souls?
How to copy the contents of all files with a certain name into a new file?
I could not break this equation. Please help me
Make it rain characters
Problems with Ubuntu mount /tmp
Why can't wing-mounted spoilers be used to steepen approaches?
The variadic template constructor of my class cannot modify my class members, why is that so?
Relations between two reciprocal partial derivatives?
Arduino Pro Micro - switch off LEDs
First use of “packing” as in carrying a gun
Install many applications using one command
How to change the limits of integration
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Integration limits of the double integral after conversion to the polar coordinatesConvergent or Divergent using LimitsHow do I solve a double integral with an absolute value?How are limits of integration changed?Converting limits of integrationHow to set the limits for Jacobian IntegrationLimits of integration in multivariable integrals during change of variablesChange of limits of integrationWhat theorem(s) is(are) used to change between the various improper integrals.how does an integral becoming negative effect limits of integration?
$begingroup$
I am attempting to solve the integral of the following...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
asked Apr 9 at 23:40
BolboaBolboa
408616
$endgroup$
add a comment |
$begingroup$
I am attempting to solve the integral of the following...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
asked Apr 9 at 23:40
BolboaBolboa
408616
$endgroup$
1
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30
add a comment |
$begingroup$
I am attempting to solve the integral of the following...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
asked Apr 9 at 23:40
BolboaBolboa
408616
$endgroup$
I am attempting to solve the integral of the following...
$$int_0^2 piint_0^inftye^-r^2rdrTheta $$
So I do the following step...
$$=2 piint_0^inftye^-r^2rdr$$
but then the next step is to substitute $s = -r^2$ which results in...
$$=2 piint_- infty^0frac12e^sds$$
The limits of integration are reversed now and the $r$ somehow results in $1/2$.
Can someone explain why this works? Why did substituting cause the limits change and result in the integration above?
calculus integration limits
calculus integration limits
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
asked Apr 9 at 23:40
BolboaBolboa
408616
408616
1
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30
add a comment |
1
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30
1
1
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181696%2fhow-to-change-the-limits-of-integration%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
add a comment |
$begingroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
add a comment |
$begingroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$s=-r^2$ gives $ds=-2rdr$ so $dr =-frac 1 2r ds$. Also, as $r$ increases from $0$ to $infty$, $s$ decreases from $0$ to $-infty$.
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Apr 9 at 23:44
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
add a comment |
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
1
1
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
$begingroup$
It should be noted that the minus sign from the substitution is then used to reverse the order of the limits.
$endgroup$
– John Doe
Apr 9 at 23:45
1
1
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
$begingroup$
@JohnDoe Right. I didn't mention it explicitly but that is what I meant.
$endgroup$
– Kavi Rama Murthy
Apr 9 at 23:49
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
$endgroup$
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
$endgroup$
add a comment |
$begingroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
$endgroup$
Do you really need substitution. We already know the antiderivative of $re^-r^2$ and it is $-e^-r^2over 2$
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
edited Apr 9 at 23:47
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
answered Apr 9 at 23:45
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,210212
2,210212
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3181696%2fhow-to-change-the-limits-of-integration%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
$$s = -r^2 implies ds = -2r dr implies -ds/2 = rdr$$ and as $r to infty$, $s to - infty$ and as $r to 0$, $s to 0$.
$endgroup$
– Mattos
Apr 9 at 23:44
$begingroup$
This constellation of consequences is why I recommend to my students to set $s = r^2$. This gives fewer minus signs and since the substituent is monotonically increasing, does not reverse the order of the bounds of integration.
$endgroup$
– Eric Towers
Apr 10 at 4:30